<span>Classify is the answer hope this helps :)</span>
Answer:
54
Explanation:
Given symbol of the element:
I⁻
Number of electrons found in an ion with the symbol:
This is a iodine ion:
For an atom of iodine:
Electrons = 53
Protons = 53
Neutrons = 74
An ion of iodine is one that has lost or gained electrons.
For this one, we have a negatively charged ion which implies that the number of electrons is 1 more than that of the protons.
So, number of electrons = 53 + 1 = 54
The number of electrons in this ion is 54
Answer:
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Explanation:
Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Explanation:
1) Data:
Water ⇒ C = 1 cal/g°C
m = 65.8 g
Ti = 31.5°C
Tf = 36.9°C
Heat, Q = ?
2) Formula:
Q = mCΔT
3) Calculations:
Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal
4) You can convert from calories to Joules using the conversion factor:
1 cal = 4.18 J
⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J
